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    Hi,
    I couldnt do the following question :

     



x = t + 2/t
     ; y =  t - \frac {2} {t}

    i) Show that  t = \frac {1}{2}(x+y)

    My working :

     t = x - \frac{2}{t}  ; t = y + \frac {2} {t}


     x - \frac{2}{t} = y + \frac {2} {t}

     x -y- \frac {4} {t} = 0

     \frac{xt-yt-4}{t} = 0

     t = \frac {4}{x-y}

    ?

    Thanks!
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    Just add the two equations together.
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    (Original post by Glutamic Acid)
    Just add the two equations together.
    OOOOOOOOOOOOOOOOOOOOOH yes!

    I feel like an idiot now!

    Thanks!
    Cheers mate!
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    The next part goes :

    ii) Hence write down the value of t corresponding to the piont (3, 1 ) on the curve

    Do u find dy/dx using  \frac {dy}{dx} = \frac {dy}{dt} X \frac {dt}{dx} ?

    This question doesnt make sense to me as :

    iii) Find the gradient of the curve at the point (3, 1) ??

    Thanks.
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    For (ii) Use the result in (i): t = \dfrac{x+y}{2}. You'll then have t.

    For (iii), you're right, find dy/dx (either by your way or \dfrac{\text{d}y}{\text{d}x} = \dfrac{\frac{\text{d}y}{\text{d}  t}}{\frac{\text{d}x}{\text{d}t}}  ).
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    (Original post by Glutamic Acid)
    For (ii) Use the result in (i): t = \dfrac{x+y}{2}. You'll then have t.

    For (iii), you're right, find dy/dx (either by your way or \dfrac{\text{d}y}{\text{d}x} = \dfrac{\frac{\text{d}y}{\text{d}  t}}{\frac{\text{d}x}{\text{d}t}}  ).

    Kooool.

    Thanks!
 
 
 
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